1. The problem statement, all variables and given/known data Find the average area of an inscribed triangle in the unit circle. Assume that each vertex of the triangle is equally likely to be at any point of the unit circle and that the location of one vertex does not affect the likelihood the location of another in any way. (Note that, as seen in Problem set 4, the maximum area is achieved by the equilateral triangle, which has side length √3 and area 3√3/4. How does the maximum compare to the average?) Hint: in order to reduce the problem to the calculation of a double integral, place one of the vertices of the triangle at(1,0),and use the polar angles θ1 and θ2 of the two other vertices as variables. What is the region of integration? Thanks http://ocw.mit.edu/NR/rdonlyres/Math...nments/ps7.pdf [Broken] (I'm not at MIT, I'm taking the class online, and this problem REALLY irritated me) 2. Relevant equations Area of a parallelogram is the cross product of two vectors, which I said were <rcosø – 1, rsinø> and <rcosß – 1, sinß>. Since we have a unit circle, I make this <cosø – 1, sinø> and <cosß – 1, sinß>; the area of the triangle formed would then be that cross product divided by 2. 3. The attempt at a solution My attempt was to say that dA = r^2 * dødß, or, in the unit circle, dødß My cross product was: sinß(cosø–1) + sinø(1–cosß). I took the absolute value of that, then called that the area, integrated it and then divided by the rectangle that the bounds would form, to get the average value of the area: 1/4π2[∫∫(sinß(1–cosø) + sinø(1–cosß))dødß] with bounds of 0 to π and 0 to π to make sure it is always positive, I changed the bounds and changed sinß(cosø–1) to sinß(1–cosø). Since the bounds would usually be 0 to 2π, I multiplied through by 22 = 4 because I would get the same result if I then made everything negative and added the same integral from π to 2π (to make sure I get absolute value) I integrated it down to the ∫ (πsinß + 2 – 2cosß) dß, and then evaluated that to get 4π. I divided that by 2, and then by π2 to get the average triangle area. My answer: ::::::::2/π:::::::: Am I on the right track or totally wrong???